If the locus of the foot of the perpendicular drawn from centre...

Slope of $OP= \dfrac{k}{h} ⇒$ slope of the tangent is $\dfrac{- h}{k}$

Hence, the equation of tangent $PQ$ is $\dfrac{( y - k )}{x - h} = \dfrac{- h}{k}$
$⇒y = - \dfrac{h}{k}x + \dfrac{h^{2} + k^{2}}{k}$
⇒ \left(\dfrac{h^{2} + k^{2}}{k}\right)2 = 40\left(\dfrac{- h}{k}\right)2 + 10\left\{$\begin{array}{ll}conditionoftangencyis \\ c^{2} = a^{2}m^{2} + b^{2}\end{array}$\right\}
$⇒ {\left(x^{2} + y^{2}\right)}^{2} = 40x^{2} + 10y^{2}$
$⇒ a – b = 30$